The average (arithmetic mean) of the positive integers x, y, and z is

Question

The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

(PS00986)

EgmatQuantExpert · Accepted Answer

Solution

Given:  
 • The average (arithmetic mean) of the positive integers x, y, and z is 3
• Also, x < y < z

To find:  
 • The greatest possible value of z

Approach and Working:   
As the average is 3, 
 • Then sum of all 3 = 3 * 3 = 9

Also, x < y < z and each are positive integers.

To maximise z, we should take minimum value of x and y
 • Minimum possible x = 1
• Minimum possible y = 2
• Therefore, maximum possible z = 9 – (1 + 2) = 6

Hence, the correct answer is option B.

Answer: B

BrentGMATPrepNow · Answer

The average (arithmetic mean) of the positive integers x, y, and z is 3  
We can write:  \frac{x+y+z}{3}=3

Multiply both sides of the equation by  3  to get:  x+y+z=9

At this point we need to find the greatest possible value of  z 
  Important  : Keep in mind that the three numbers are DIFFERENT POSITIVE INTEGERS

Since we know that  x+y+z=9 , we can MAXIMIZE the value of  z  by MINIMIZING the values of  x  and  y 
We know that  x  is the smallest value. 
Since  x  must be a positive integer, the smallest possible value of  x  is  1  
Since  y  must be different from  x , the smallest possible value of  y  is  2  
At this point we have maximized the value of  z

If  x=1  and  y=2 , then  z=6

Answer: B

Cheers,
Brent

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Shbm · Answer

Bunuel  so zero isn't counted as an integer? This always confuses me.

Bunuel · Answer

0 is an integer but it's neither positive nor negative.

ZERO.

1. 0 is an integer.

2. 0 is an even integer. An even number is an   integer   that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check for more below threads:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

gerikso · Answer

The smallest positive integer is 1 and 2nd smallest 2.
So, 3*3 = 3+x.
Hence, x= 6. B is the correct choice.

Why does everyone automatically assume that X and Y are different integers? I know that OG solution says that, but here every solution implies this by default. Is this a rule for this type of question - assume X and Y are different unless specified otherwise?

ScottTargetTestPrep · Answer

We can create the equation for the average of the three numbers:

(x + y + z)/3 = 3

x + y + z = 9

Since z is the largest integer and we want it to be as large as possible, we can let x = 1 and y = 2, and thus the largest value z could be is 6.

Answer: B

ayoubb · Answer

Why can't x and y be equal? the question does not say different integers, that way x=y=1 and Z=7 ?  Bunuel  could you please clarify

Bunuel · Answer

I guess you missed the highlighted part in the question:

The average (arithmetic mean) of the positive integers x, y, and z is 3. If  x < y < z , what is the greatest possible value of z ?

jaydavido · Answer

can someone explain why zero isn't correct? is it because its neither positive or negative integer?

Bunuel · Answer

Notice that the question says that x, y, and z are positive integers, while 0 is neither positive nor negative integer.

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer.

3. Zero is  neither positive nor negative  (the only one of this kind).

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{x}{0} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x).

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

ArnauG · Answer

To find the greatest possible value of z, we need to maximize the value of z while satisfying the given conditions.

Given that the average of x, y, and z is 3, we can express this as:

(x + y + z) / 3 = 3.

Multiplying both sides of the equation by 3, we have:

x + y + z = 9.

Since x < y < z, and we want to maximize z, we can start by assigning the minimum possible values to x and y, which are 1 and 2, respectively.

Plugging these values into the equation, we have:

1 + 2 + z = 9,

which simplifies to:

3 + z = 9.

Solving for z, we find:

z = 6.

Therefore, the greatest possible value of z is 6.

Hence, the correct answer is (B) 6.

Budhaditya_Saha · Answer

Bunuel   EgmatQuantExpert  According to this solution, the set is {1,2,6}. Now we can get the mean by adding the min and max values of the set and then dividing it by 2. So in this case, (1+6)/2 = 3.5, but the given mean is 3. So hw come this answer is right?

If the set would have been {1,3,5} then the mean is 3 by both (1+3+5)/3 & (1+5)/2. So shouldnt '5' be the greatest value of z?

Bunuel · Answer

The average (arithmetic mean) is calculated by dividing the sum of the values by the number of values.

So, the average of 1, 2, and 6 is (1 + 2 + 6)/3 = 9/3 = 3.

The method you describe, finding the average by averaging the smallest and largest terms, works for sets that are evenly spaced. For example, the average of 1, 3, 5, 7 is (1 + 3 + 5 + 7)/4 = 4, which also equals (1 + 7)/2 = 4.

In the given problem, since 1, 2, and 6 are not evenly spaced, this method does not apply.

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The average (arithmetic mean) of the positive integers x, y, and z is

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